[/quote]
,
and 
be positive real numbers such that 
.
with their incenter 
.
lies on 
on the same side as 
wrt 
.
lies on the shorter arc 
of the circumcircle 
Prove that 
is the angle bisector of 
.
Prove that
be the set of all sequences 
of non-negative integers such that
;
such that 
for all 
.
to be the sequence 
,
,
is the number of 
'
.
,
and 
.
satisfies the following property:![\[
f\left( \lambda x + (1 - \lambda)x',\; \lambda y + (1 - \lambda)y' \right) > \min\{f(x, y),\; f(x', y')\}
\]](http://latex.artofproblemsolving.com/3/b/d/3bd9fbf82b0b1d98a3c68745bb3f159cd9e2529a.png)
and for all 
.
be a subset of 
.
such that for any integer 
,
so that 
and all the digits in the decimal representations of 
.
Let 
be such that any positive integer 
can be written as 
where the non-negative integers 
have all their digits in 
Find the smallest possible number of elements in 
,
denote the number of elements in the set 
be the set of all functions![\[ f : \{1, 2, \ldots, n\} \to \{1, 2, \ldots, k\} \quad \text{with } n \geq 3,\; k \geq 2 \]](http://latex.artofproblemsolving.com/b/1/b/b1bf08934378882b9cd0b7d983f83b15453acca9.png)
satisfying the condition:![\[ f(i) \ne f(i+1) \quad \text{for every } i,\; 1 \leq i \leq n-1. \]](http://latex.artofproblemsolving.com/f/d/1/fd11b44d7fc482b5c6018e19161f6e30b7835c2d.png)
Show that![\[ |F| = k(k-1)^{n-1}. \]](http://latex.artofproblemsolving.com/a/8/8/a88a07984bad88a854de4dad8c24424aaad38365.png)
denote the number of functions in 
.
,![\[ c(n, k) = k(k-1)^{n-1} - c(n-1, k). \]](http://latex.artofproblemsolving.com/8/7/2/87285a27401b9853e72b45249196954082d9ec30.png)
,![\[ c(n, k) = (k-1)^n - (-1)^n(k-1). \]](http://latex.artofproblemsolving.com/2/f/c/2fcbcb385fcc1b3956232dadfd7ec4a31f4d71d6.png)
,
be the orthocenter of it and 
be any point on the side 
.
are on the segments 
,
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
.
and 
lies on side 
lies on side 
.
and 
be points on segments 
and 
,
is parallel to 
be a point on line 
,
.
be the point on line 
,
.
,
-
and 
such that 
is divisible by 
?
such that for all 
,![\[f(x+yf(x))+y = xy + f(x+y).\]](http://latex.artofproblemsolving.com/c/f/3/cf3d20a041c27244e90876119b4568b7a3e13c03.png)
is orthocenter, 
is the midpoint of 
is the midpoint of smaller arc 
is on 
such that, 
is perpendicular to 
.
is paralelogram(Synthetic sol needed).
Y by samrocksnature, Adventure10, ImSh95, Mango247, and 1 other user
Let 
be an increasing sequence of nonnegative integers such that every nonnegative integer can be expressed uniquely in the form 
, where 
and 
are not necessarily distinct. Determine 
.
be an increasing sequence of nonnegative integers such that every nonnegative integer can be expressed uniquely in the form 
, where 
and 
are not necessarily distinct. Determine 
.Attachments:
This post has been edited 1 time. Last edited by orl, Oct 23, 2004, 12:48 PM
.
can be represented in the mentioned way, we must have 
(this will be important a bit later). We have 
.
and we get 
.
.
,
,
we have 
,
,
.
'
.
.
is the sum of the first 
numbers. So 
.
.
is the sum of the first 
is simply the result of writing 
and adding it on to the current sequence.
for some integer 
.
.
,
has exactly 
,
equals 
.
in the desired form.
digits, possibly with leading 0s, so we have 
.
in base 8, 
in base 8, and 
in base 8.
with 
,
possible numbers of the form 
and 
for any integer 
,
.
as 
and observe that this formal series is defined at all values 
of the variable. The given relation is equivalent to 
Replacing 
reveals that 
and we conclude that the formal series 
is given by 
Thus, 
when written in base 
where 
,
.![\[a_n = \sum_{i\geq 0} e_i\cdot 8^{i}\]](http://latex.artofproblemsolving.com/8/1/1/811496f4572bcdc41c162211ddb8698e91595004.png)
,
,
for 
.
,
![\[1998 = 11111001110_2 \Longrightarrow a_{1998}=11111001110_8 = 1227096648_{10}\]](http://latex.artofproblemsolving.com/a/2/f/a2f0971eac26d469933ffd73e7f02c425c2c1653.png)
and we are done.
contains precisely the nonnegative integers that contain no digits other than 0 and 1 in base 8. Clearly, this sequence works, since for each nonnegative integer, we essentialy express it in base 8, split each digit into 3 bits, and use those to pick either 0 or 1 in that corresponding location in order to find the bit in that place for 
.
,
denote the smallest nonnegative integer that cannot be expressed as 
.
.
,
as 
However, we can also express 
,
,
for all 
However, consider 
through 
for 
,
.
.
On the other hand, combining this with 
,
,
and thus 
can be characterized as the sequence of numbers expressible as the sum of powers of 
,
,
,
.
,
and the third being every digit of place value 
and the number formed only using digit first class is 
,
and the third class forms 
.
t
terms. We claim by induction that the first 
integers.
.
possibilities are expressed 
we simply need those 
up to 
.
.
.
so 
so 
.
is a term in 
,
is a term in 
,
must both be terms in 
,
.
and 
,
bits determine 
bits determine 
bits determine 
,
.
,![\[f(x)f(x^2)f(x^4) = 1 + x + x^2 + \dots = \dfrac{1}{1 - x} \implies \dfrac{f(x)}{f(x^8)} = \dfrac{f(x)f(x^2)f(x^4)}{f(x^2)f(x^4)f(x^8)} = \dfrac{1 - x^2}{1 - x} = 1 + x.\]](http://latex.artofproblemsolving.com/9/4/f/94f8f5bd31d28d7dbb6953d923ac0b401e973b98.png)
Therefore, we have the telescoping series![\[f(x) = \dfrac{f(x)}{f(x^8)} \cdot \dfrac{f(x^8)}{f(x^{8^2})} \cdot \dfrac{f(x^{8^2})}{f(x^{8^3})} \cdots = (1 + x)(1 + x^8)(1 + x^{8^2}) \cdots \]](http://latex.artofproblemsolving.com/7/4/d/74d164c9ffcff327b0696379c49cd65471b8eab1.png)
So ![\[1998 = 11111001110_2 \implies a_{1998} = 8^{10} + 8^9 + 8^8 + 8^7 + 8^6 + 8^3 + 8^2 + 8^1\]](http://latex.artofproblemsolving.com/4/2/7/427f6c438222c5a01596beefc19e669bfb242710.png)
as wanted.
and 
with 
.
(if not we can't get 1). As 
can be achieved by using 1 and 0, but not 8, 
and 
.
uniquely. Write 
,
and 
,
.
,
,
,
,
,
,
.
.
,
(because 
)
=
.
,
just using 
by induction hypothesis. Therefore, we select those exact indices for 
instead of 
.
just using 
,
,
for 
.
and as 
,
and 
as otherwise we would be unable to represent 
The following claim determines that at most one sequence works:
for 
If 
then we could represent the number 
then we can never form 
as we always use a larger number. This concludes our proof of the claim.
where 
is the binary representation of 
This works because the only way to account for the 
positions in the binary representation are with the 
and 
positions. This uniquely determines the 
Hence this is the only sequence which works. Finally, 
so our answer is 
and then eventually we reach a number not representable, in which we have to add that number to the sequence. This process is deterministic.
inclusive (uniquely) all numbers can be represented uniquely by considering each digit separately. Therefore 
Hence the answer is 
